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# 1352.Product-of-the-Last-K-Numbers

## 题目描述

Implement the class ProductOfNumbers that supports two methods:
Adds the number num to the back of the current list of numbers.
2. getProduct(int k)
Returns the product of the last k numbers in the current list.
You can assume that always the current list has at least k numbers.
At any time, the product of any contiguous sequence of numbers will fit into a single 32-bit integer without overflowing.
Example:
Input
[[],[3],[0],[2],[5],[4],[2],[3],[4],[8],[2]]
Output
[null,null,null,null,null,null,20,40,0,null,32]
Explanation
ProductOfNumbers productOfNumbers = new ProductOfNumbers();
productOfNumbers.getProduct(2); // return 20. The product of the last 2 numbers is 5 * 4 = 20
productOfNumbers.getProduct(3); // return 40. The product of the last 3 numbers is 2 * 5 * 4 = 40
productOfNumbers.getProduct(4); // return 0. The product of the last 4 numbers is 0 * 2 * 5 * 4 = 0
productOfNumbers.getProduct(2); // return 32. The product of the last 2 numbers is 4 * 8 = 32
Constraints:
There will be at most 40000 operations considering both add and getProduct.
0 <= num <= 100
1 <= k <= 40000

## 代码

### Approach #1

class ProductOfNumbers {
List<Integer> prod;
int p = 1;
public ProductOfNumbers() {
prod = new ArrayList();
}
public void add(int num) {
if (num == 0) {
p = 1;
prod = new ArrayList();
} else {
p *= num;
}
}
public int getProduct(int k) {
if (k >= prod.size()) {
return 0;
}
int last = prod.get(prod.size() - 1);
int prev = prod.get(prod.size() - 1 - k);
return last / prev;
}
}
/**
* Your ProductOfNumbers object will be instantiated and called as such:
* ProductOfNumbers obj = new ProductOfNumbers();